So, given an integral domain R, it is often very useful to know that R has a Euclidean function: in particular, this implies that R is a PID. However, if there is no "obvious" Euclidean function, then determining whether R is a PID is generally a much easier problem than determining whether it is a Euclidean domain.

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Let R be an integral domain. A Euclidean function on R is a function f{\displaystyle f} from R∖{0}{\displaystyle R\setminus \{0\}} to the non-negative integers satisfying the following fundamental division-with-remainder property:

(EF1) If a and b are in R and b is nonzero, then there are q and r in R such that a = bq + r and either r = 0 or f(r) < f(b).

A Euclidean domain is an integral domain which can be endowed with at least one Euclidean function. It is important to note that a particular Euclidean function f is not part of the structure of a Euclidean domain: in general, a Euclidean domain will admit many different Euclidean functions.

Most algebra texts require a Euclidean function to have the following additional property:

(EF2) For all nonzero a and b in R, f(a) ≤ f(ab).

However, one can show that (EF2) is superfluous in the following sense: any domain R which can be endowed with a function g satisfying (EF1) can also be endowed with a function f satisfying (EF1) and (EF2): indeed, for a∈R∖{0}{\displaystyle \scriptstyle a\in R\setminus \{0\}} one can define f(a) as follows[1]

Many authors use other terms such as "degree function", "valuation function", "gauge function" or "norm function", in place of "Euclidean function".[citation needed] Some authors also require the domain of the Euclidean function to be the entire ring R;[2] however this does not essentially affect the definition, since (EF1) does not involve the value of f(0). The definition is sometimes generalized by allowing the Euclidean function to take its values in any well-ordered set; this weakening does not affect the most important implications of the Euclidean property.

The property (EF1) can be restated as follows: for any principal ideal I of R with nonzero generator b, all nonzero classes of the quotient ring R/I have a representative r with f(r) < f(b). Since the possible values of f are well-ordered, this property can be established by proving f(r) < f(b) for any r (not in I) with minimal value of f(r) in its class. Note that for a Euclidean function that is so established there need not exist an effective method to determine q and r in (EF1).

K[[X]], the ring of formal power series over the field K. For each nonzero power series P, define f(P) as the degree of the smallest power of X occurring in P. In particular, for two nonzero power series P and Q, f(P)≤f(Q) iff P divides Q.

Any discrete valuation ring. Define f(x) to be the highest power of the maximal ideal M containing x (equivalently, to the power of the generator of the maximal ideal that x is associated to). The previous case K[[X]] is a special case of this.

Every domain that is not a principal ideal domain, such as the ring of polynomials in at least two indeterminates over a field, or the ring of univariate polynomials with integer coefficients

The ring of integers of Q(−19),{\displaystyle \mathbb {Q} ({\sqrt {-19}}),} consisting of the numbers a+b−192{\displaystyle {\frac {a+b{\sqrt {-19}}}{2}}} such that a and b are integers, that are either both even or both odd. It is a principal ideal domain that is not Euclidean.

The ring R[X,Y]/(X2+Y2+1){\displaystyle \mathbb {R} [X,Y]/(X^{2}+Y^{2}+1)} is also a principal ideal domain that is not Euclidean.[citation needed]

R is a principal ideal domain (PID). In fact, if I is a nonzero ideal of R then any element a of I\{0} with minimal value (on that set) of f(a) is a generator of I.[5] As a consequence R is also a unique factorization domain and a Noetherian ring. With respect to general principal ideal domains, the existence of factorizations (i.e., that R is an atomic domain) is particularly easy to prove in Euclidean domains: choosing a Euclidean function f satisfying (EF2), x cannot have any decomposition into more than f(x) nonunit factors, so starting with x and repeatedly decomposing reducible factors is bound to produce a factorization into irreducible elements.

Any element of R at which f takes its globally minimal value is invertible in R. If an f satisfying (EF2) is chosen, then the converse also holds, and f takes its minimal value exactly at the invertible elements of R.

If the Euclidean property is algorithmic, i.e., if there is a division algorithm that for given a and nonzero b produces a quotient q and remainder r with a = bq + r and either r = 0 or f(r) < f(b), then an extended Euclidean algorithm can be defined in terms of this division operation.[6]

If a Euclidean domain is not a field then it has an element a with the following property: any element x not divisible by a can be written as x=ay+u for some unit u and some element y. This follows by taking a to be a non-unit with f(a) as small as possible. This strange property can be used to show that some principal ideal domains are not Euclidean domains, as not all PIDs have this property. For example, for d = −19, −43, −67, −163, the ring of integers of Q(d){\displaystyle \mathbb {Q} ({\sqrt {d}})} is a PID which is not Euclidean, but the cases d = −1, −2, −3, −7, −11 are Euclidean.[7]

However, in many finite extensions of Q with trivial class group, the ring of integers is Euclidean (not necessarily with respect to the absolute value of the field norm; see below). Assuming the extended Riemann hypothesis, if K is a finite extension of Q and the ring of integers of K is a PID with an infinite number of units, then the ring of integers is Euclidean.[8] In particular this applies to the case of totally real quadratic number fields with trivial class group. In addition (and without assuming ERH), if the field K is a Galois extension of Q, has trivial class group and unit rank strictly greater than three, then the ring of integers is Euclidean.[9] An immediate corollary of this is that if the number field is Galois over Q, its class group is trivial and the extension has degree greater than 8 then the ring of integers is necessarily Euclidean.

Algebraic number fieldsK come with a canonical norm function on them: the absolute value of the field normN that takes an algebraic element α to the product of all the conjugates of α. This norm maps the ring of integers of a number field K, say OK, to the nonnegative rational integers, so it is a candidate to be a Euclidean norm on this ring. If this norm satisfies the axioms of a Euclidean function then the number field K is called norm-Euclidean or simply Euclidean.[10][11] Strictly speaking it is the ring of integers that is Euclidean since fields are trivially Euclidean domains, but the terminology is standard.

If a field is not norm-Euclidean then that does not mean the ring of integers is not Euclidean, just that the field norm does not satisfy the axioms of a Euclidean function. In fact, the rings of integers of number fields may be divided in several classes:

Those that are not principal and therefore not Euclidean, such as the integers of Q(−5){\displaystyle \mathbb {Q} ({\sqrt {-5}})}

Those that are principal and not Euclidean, such as the integers of Q(−19){\displaystyle \mathbb {Q} ({\sqrt {-19}})}

Those that are Euclidean and not norm-Euclidean, such as the integers of Q(69){\displaystyle \mathbb {Q} ({\sqrt {69}})}[12]

Those that are norm-Euclidean, such as Gaussian integers (integers of Q(−1){\displaystyle \mathbb {Q} ({\sqrt {-1}})})

The norm-Euclidean quadratic fields have been fully classified, they are Q(d){\displaystyle \mathbb {Q} ({\sqrt {d}})} where d takes the values